HINTS & SOLUTIONS

Answer: (a) Cm
Explanation: Electric dipole moment is defined as $p = q \times 2a$, where $q$ is charge (measured in Coulombs, C) and $2a$ is distance (measured in meters, m). Hence, the SI unit is Coulomb-meter (C m).

Answer: (d) 28 A
Solution: Electric current $I = \frac{dq}{dt}$.
$I = \frac{d}{dt}(5t^{2} + 8t) = 10t + 8$.
At $t = 2\text{ s}$, $I = 10(2) + 8 = 20 + 8 = 28 \text{ A}$.

Answer: (c) straight line
Explanation: Magnetic force $\vec{F} = q(\vec{v} \times \vec{B}) = qvB\sin\theta$. If the velocity is parallel to the magnetic field, $\theta = 0^\circ$ and $\sin 0^\circ = 0$. Since no magnetic force acts on it, it continues to move in a straight line.

Answer: (a) $\frac{1}{\sqrt{2}} A$
Solution: Power $P = V_{rms} I_{rms}$.
$I_{rms} = \frac{P}{V_{rms}} = \frac{12}{24} = 0.5 \text{ A} = \frac{1}{2} \text{ A}$.
Peak current $I_{0} = \sqrt{2} \times I_{rms} = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}} \text{ A}$.

Answer: (c) accelerating charge
Explanation: A static charge produces only an electric field. A charge moving with uniform velocity produces steady electric and magnetic fields. Only an accelerating (or oscillating) charge produces changing electric and magnetic fields that propagate as electromagnetic waves.

Answer: (b) may see a primary and a secondary rainbow as concentric circles
Explanation: A rainbow is actually a full circle of light centered on the antisolar point. From the ground, the lower half is blocked by the Earth's horizon. From an airplane high above the ground, the entire concentric circular arcs of primary and secondary rainbows can be visible.

Answer: (a) $\frac{h\nu}{c}$
Explanation: The momentum of a photon is given by de-Broglie's relation $p = \frac{h}{\lambda}$. Since $c = \nu\lambda \implies \lambda = \frac{c}{\nu}$, substituting this yields $p = \frac{h\nu}{c}$.

Answer: (b) ${}_{26}^{56}Fe$
Explanation: Iron-56 (${}_{26}^{56}Fe$) has the highest binding energy per nucleon (approximately 8.8 MeV) among all nuclides, making it the most stable nucleus in nature.

Answer: (c) $I_{e}=I_{b}+I_{c}$
Explanation: According to Kirchhoff's current law applied to a transistor, the total current leaving the emitter is the sum of the current entering the base and the current entering the collector (or vice versa). Thus, $I_e = I_b + I_c$ always holds true.

Answer: (b) absorption and scattering
Explanation: Attenuation is the loss of optical power as light travels through the fiber. It is primarily caused by the absorption of light by impurities in the glass and Rayleigh scattering due to microscopic variations in the material density.

Answer: (c) Interference of light
Explanation: The brilliant iridescent colors seen in peacock feathers and some insect wings are structural colors caused by the interference of light reflecting off complex, microscopic layered structures within the feathers (thin-film interference).

Answer: (b) Infinite
Explanation: A plane mirror can be considered a spherical mirror with an infinitely large radius. Because it has no curvature, its focal length and radius of curvature are both infinite ($R = \infty$).

Solution:
Electric field lines represent the direction of the net electric field at any point. If two lines were to cross each other, it would mean that at the point of intersection, there are two different directions for the electric field (represented by drawing two tangents). Since a net electric field at a single point can only have one unique direction, two lines can never intersect.

Solution:
The opposition offered by a capacitor to the flow of current is called capacitive reactance, given by $X_{C} = \frac{1}{2\pi f C}$, where $f$ is frequency.
For direct current (D.C.), the frequency $f = 0$. Substituting this into the formula gives $X_{C} = \frac{1}{0} = \infty$. Thus, a capacitor offers infinite resistance to D.C., effectively blocking it.

Solution:
Average electric energy density $u_{E} = \frac{1}{2}\epsilon_{0}E^{2}_{rms}$.
In an EM wave, $E_{rms} = cB_{rms}$ and the speed of light $c = \frac{1}{\sqrt{\mu_{0}\epsilon_{0}}}$.
Substituting $E_{rms}$: $u_{E} = \frac{1}{2}\epsilon_{0}(cB_{rms})^{2} = \frac{1}{2}\epsilon_{0}c^{2}B^{2}_{rms}$.
Substituting $c^{2}$: $u_{E} = \frac{1}{2}\epsilon_{0} \left(\frac{1}{\mu_{0}\epsilon_{0}}\right) B^{2}_{rms} = \frac{B^{2}_{rms}}{2\mu_{0}}$.
Since $\frac{B^{2}_{rms}}{2\mu_{0}}$ is the average magnetic energy density ($u_{B}$), we have shown $u_{E} = u_{B}$. Solution (Alternative Numerical):
Using the relation $\lambda = \frac{c}{\nu}$ where $c = 3\times10^{8} \text{ m/s}$.
For $\nu_{1} = 7.5 \text{ MHz} = 7.5 \times 10^{6} \text{ Hz}$: $\lambda_{1} = \frac{3\times10^{8}}{7.5\times10^{6}} = 40 \text{ m}$.
For $\nu_{2} = 12 \text{ MHz} = 12 \times 10^{6} \text{ Hz}$: $\lambda_{2} = \frac{3\times10^{8}}{12\times10^{6}} = 25 \text{ m}$.
The corresponding wavelength band is 25 m to 40 m.

Solution:
Einstein's Photoelectric Equation:
According to Planck's quantum theory, light consists of energy packets called photons, each with energy $E = h\nu$.
When a photon hits a metal surface, its energy is used in two ways:
1. A part of the energy is used to free the electron from the surface barrier. This minimum energy is the work function ($\Phi_{0} = h\nu_{0}$).
2. The remaining energy provides kinetic energy ($K_{max}$) to the emitted photoelectron.
By conservation of energy: $E = \Phi_{0} + K_{max}$
$h\nu = h\nu_{0} + \frac{1}{2}mv^{2}_{max} \implies K_{max} = h(\nu - \nu_{0})$.

Solution:
NOR Gate:
Boolean Expression: $Y = \overline{A + B}$
Logic Symbol: An OR gate symbol with a small bubble (inversion) at the output.
Truth Table:

A	B	Y
0	0	1
0	1	0
1	0	0
1	1	0

Solution (Alternative):
p-n junction diode: A semiconductor device formed by joining a p-type semiconductor and an n-type semiconductor. It allows current to flow easily in forward bias but restricts it in reverse bias.
Characteristic Curve: A graph plotting Forward Current (mA) vs Forward Voltage (+V) in the first quadrant, showing an exponential rise after the "knee voltage". In the third quadrant, it plots Reverse Current ($\mu$A) vs Reverse Voltage (-V), showing a tiny constant saturation current until breakdown.

Solution:
Total Internal Reflection (TIR): The phenomenon where a light ray traveling from a denser medium to a rarer medium is completely reflected back into the denser medium if it strikes the interface at an angle greater than the critical angle.
Conditions for TIR: 1. Light must travel from an optically denser medium to an optically rarer medium. 2. The angle of incidence in the denser medium must be greater than the critical angle ($i > i_{c}$) for that pair of media. Solution (Alternative):
Blue Colour of Sky: As sunlight passes through the atmosphere, it undergoes Rayleigh scattering by air molecules. According to Rayleigh's law, the intensity of scattered light is inversely proportional to the fourth power of wavelength ($I \propto \frac{1}{\lambda^{4}}$). Since blue light has a much shorter wavelength than red light, it gets scattered much more strongly in all directions, illuminating the sky with a blue tint.

Solution:
Internal resistance using Potentiometer:
Let a cell of emf $E$ and internal resistance $r$ be connected to the potentiometer. A resistance box $R$ is connected in parallel with the cell through a key $K$.
Step 1: When key $K$ is open, no current is drawn from the cell. The balance point length $l_{1}$ gives the emf of the cell: $E \propto l_{1} \implies E = k l_{1}$.
Step 2: When key $K$ is closed, current is drawn through $R$. The terminal potential difference $V$ is balanced at a new length $l_{2}$: $V \propto l_{2} \implies V = k l_{2}$.
We know internal resistance $r = \left(\frac{E}{V} - 1\right) R$.
Substituting the values: $r = \left(\frac{k l_{1}}{k l_{2}} - 1\right) R \implies r = \left(\frac{l_{1}}{l_{2}} - 1\right) R$.

Solution:
SI unit of Electric energy: Joule (J). (Commercial unit is kWh).
Numerical: Given Resistance $R = 25 \text{ }\Omega$, Voltage $V = 220 \text{ V}$, Time $t = 1 \text{ minute} = 60 \text{ s}$.
Heat produced is given by Joule's Law of Heating: $H = \frac{V^{2}}{R} t$.
$H = \frac{(220)^{2}}{25} \times 60$
$H = \frac{48400}{25} \times 60 = 1936 \times 60$
$H = 116160 \text{ Joules}$ (or $116.16 \text{ kJ}$).

Solution:
Ampere's Circuital Law: It states that the line integral of the magnetic field $\vec{B}$ around any closed loop is equal to $\mu_{0}$ times the total steady current $I$ passing through the surface enclosed by the loop. $\oint \vec{B} \cdot d\vec{l} = \mu_{0} I$.
Proof: Consider a long straight wire carrying current $I$. Magnetic field lines are concentric circles. Choose a circular Amperian loop of radius $r$.
At any point on the loop, $B = \frac{\mu_{0}I}{2\pi r}$. The vectors $\vec{B}$ and $d\vec{l}$ are parallel ($\theta = 0^\circ$).
L.H.S = $\oint \vec{B} \cdot d\vec{l} = \oint B dl \cos(0) = B \oint dl$.
Since $\oint dl = 2\pi r$, L.H.S = $\left(\frac{\mu_{0}I}{2\pi r}\right) \times 2\pi r = \mu_{0} I$.
Thus, L.H.S = R.H.S. Hence proved. Solution (Alternative):
Potential Energy of Bar Magnet:
When a bar magnet of dipole moment $M$ is placed in a uniform magnetic field $B$ at an angle $\theta$, it experiences a restoring torque $\tau = MB \sin\theta$.
Work done to rotate it by $d\theta$ is $dW = \tau d\theta = MB \sin\theta d\theta$.
Total work done to rotate it from $\theta_1$ to $\theta_2$ is $W = \int_{\theta_1}^{\theta_2} MB \sin\theta d\theta = MB [-\cos\theta]_{\theta_1}^{\theta_2} = -MB(\cos\theta_2 - \cos\theta_1)$.
Assuming potential energy is zero at $\theta_1 = 90^\circ$, the potential energy at angle $\theta$ is $U = -MB\cos\theta = -\vec{M} \cdot \vec{B}$.
Maximum: At $\theta = 180^\circ$ (anti-parallel), $U = -MB(-1) = +MB$ (Unstable equilibrium).
Minimum: At $\theta = 0^\circ$ (parallel), $U = -MB(1) = -MB$ (Stable equilibrium).

Solution:
Force on a straight conductor:
Consider a conductor of length $l$ and cross-sectional area $A$ carrying current $I$ in a uniform magnetic field $\vec{B}$.
Let $n$ be the number density of free electrons. Total number of electrons $N = nAl$.
Force on one electron drifting with velocity $v_{d}$ is $\vec{f} = -e(\vec{v}_{d} \times \vec{B})$.
Total magnetic force $\vec{F} = N\vec{f} = nAl [-e(\vec{v}_{d} \times \vec{B})] = -(neAv_{d})l \times \vec{B}$.
Since current $I = neAv_{d}$ and defining current element vector $I\vec{l}$ opposite to electron drift, we get $\vec{F} = I(\vec{l} \times \vec{B})$.
Magnitude of force $F = IlB\sin\theta$.
Maximum Force: Force is maximum when the conductor is placed perpendicular to the magnetic field ($\theta = 90^\circ$). $\sin 90^\circ = 1$.
$F_{max} = IlB$.

Solution:
Width of Central Maximum vs Secondary Maximum:
In single slit diffraction, the slit width is $a$. Let $D$ be the distance to the screen.
The angular position of the $n$-th minimum is given by $\theta_{n} = n\frac{\lambda}{a}$.
The central maximum lies between the first minima on either side ($n = +1$ and $n = -1$).
Angular width of central maximum $= \theta_{+1} - \theta_{-1} = \frac{\lambda}{a} - \left(-\frac{\lambda}{a}\right) = \frac{2\lambda}{a}$.
Linear width of central maximum $W_c = D \times (\text{Angular width}) = \frac{2\lambda D}{a}$.
A secondary maximum (e.g., the first one) lies between the 1st minimum and 2nd minimum.
Angular width of secondary maximum $= \theta_2 - \theta_1 = \frac{2\lambda}{a} - \frac{\lambda}{a} = \frac{\lambda}{a}$.
Linear width of secondary maximum $W_s = D \times \frac{\lambda}{a} = \frac{\lambda D}{a}$.
Comparing the two: $W_c = 2 \times W_s$. Hence proved.

Solution:
Laws of Radioactive Decay:
1. Radioactivity is a spontaneous, random nuclear phenomenon independent of external physical conditions.
2. A nucleus can emit an alpha or beta particle, but not both simultaneously.
3. Decay Law: The rate of disintegration of radioactive nuclei is directly proportional to the number of undecayed nuclei present at that instant. $-\frac{dN}{dt} \propto N \implies \frac{dN}{dt} = -\lambda N$, where $\lambda$ is the decay constant. Integrating yields $N = N_0 e^{-\lambda t}$. Solution (Alternative Numerical):
Nitrogen nucleus ${}_{7}^{14}N$ has $Z=7$ protons and $A-Z = 14-7 = 7$ neutrons.
Mass of 7 protons $= 7 \times 1.007825 = 7.054775 \text{ amu}$.
Mass of 7 neutrons $= 7 \times 1.008665 = 7.060655 \text{ amu}$.
Sum of constituent masses $= 7.054775 + 7.060655 = 14.115430 \text{ amu}$.
Actual nuclear mass $= 14.00307 \text{ amu}$.
Mass defect $\Delta m = 14.115430 - 14.00307 = 0.11236 \text{ amu}$.
Binding Energy (BE) $= \Delta m \times 931.5 \text{ MeV} = 0.11236 \times 931.5 = 104.66 \text{ MeV}$.
Binding Energy per nucleon $= \frac{\text{BE}}{A} = \frac{104.66}{14} = 7.47 \text{ MeV/nucleon}$.

Solution:
NPN Transistor as CE Amplifier:
In Common Emitter (CE) configuration, the emitter is grounded. The emitter-base junction is forward-biased, and the collector-base junction is reverse-biased.
Working: A weak alternating input signal is applied across the base-emitter junction. During the positive half cycle, the forward bias increases, causing base current ($I_B$) to increase. This causes a massive increase in collector current ($I_C = \beta I_B$). This increases the voltage drop across the output load resistor $R_L$.
Output voltage $V_{CE} = V_{CC} - I_C R_L$. As $I_C R_L$ increases, $V_{CE}$ drops. Thus, a positive input results in a negative amplified output ($180^\circ$ phase shift). Solution (Alternative):
Zener Diode as Voltage Regulator:
A Zener diode is a heavily doped p-n junction designed to operate in the reverse breakdown region without damage.
Working: It is connected in reverse bias parallel to the load across a fluctuating DC supply, with a series dropping resistor ($R_s$). Once the input voltage exceeds the Zener breakdown voltage ($V_Z$), the diode conducts heavily. Any further increase in input voltage simply causes more current to flow through the Zener diode and $R_s$, increasing the voltage drop across $R_s$ but leaving the voltage across the Zener diode (and the parallel load) perfectly constant at $V_Z$.

Solution:
Communication System: A system set up to transmit information from one location to another reliably over a distance.
Block Diagram flow:
Information Source $\rightarrow$ Message Signal $\rightarrow$ Transmitter $\rightarrow$ Transmitted Signal $\rightarrow$ Communication Channel (with Noise added) $\rightarrow$ Received Signal $\rightarrow$ Receiver $\rightarrow$ Message Signal $\rightarrow$ User of Information.

Solution:
Electric Potential: The amount of work done in bringing a unit positive test charge from infinity to that point against the electrostatic force, without acceleration. $V = \frac{W}{q}$.
Derivation:
Consider a point charge $+q$ at the origin. Let P be a point at distance $r$.
Let a test charge $+q_{0}$ be placed at distance $x$ from $+q$. Coulomb force $F = \frac{1}{4\pi\epsilon_{0}}\frac{qq_{0}}{x^{2}}$ (repulsive outward).
Work done to move $q_{0}$ by a small distance $dx$ towards $q$ is $dW = \vec{F} \cdot d\vec{x} = F dx \cos(180^\circ) = -F dx$.
Total work $W$ from $x=\infty$ to $x=r$ is:
$W = \int_{\infty}^{r} -\frac{1}{4\pi\epsilon_{0}}\frac{qq_{0}}{x^{2}} dx = -\frac{qq_{0}}{4\pi\epsilon_{0}} \left[ -\frac{1}{x} \right]_{\infty}^{r} = \frac{qq_{0}}{4\pi\epsilon_{0}r}$.
Electric potential $V = \frac{W}{q_{0}} = \frac{1}{4\pi\epsilon_{0}}\frac{q}{r}$. Solution (Alternative Numerical):
Coulomb's Law: The electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.
Vector form: $\vec{F}_{12} = \frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{r^{2}} \hat{r}_{21}$.
Calculation: For $\alpha$-particles, charge $q_{1} = q_{2} = +2e = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \text{ C}$. Distance $r = 3.2 \times 10^{-15} \text{ m}$.
$F = \frac{1}{4\pi\epsilon_{0}} \frac{q_{1}q_{2}}{r^{2}} = (9 \times 10^{9}) \times \frac{(3.2 \times 10^{-19}) \times (3.2 \times 10^{-19})}{(3.2 \times 10^{-15})^{2}}$
$F = 9 \times 10^{9} \times \frac{(3.2)^{2} \times 10^{-38}}{(3.2)^{2} \times 10^{-30}} = 9 \times 10^{9} \times 10^{-8} = 90 \text{ N}$.

Solution:
Transformer Principle: Works on Mutual Induction. When a changing alternating current flows through a primary coil, it induces an alternating EMF of the same frequency in a neighboring secondary coil.
Construction: Two coils (Primary $N_p$ and Secondary $N_s$) of insulated copper wire are wound on a closed, laminated soft iron core. The lamination minimizes eddy current losses.
Theory: Let an alternating voltage $V_p$ be applied to the primary. It sets up an alternating flux $\phi$ in the core. Assuming ideal coupling, the same flux links each turn of both coils.
By Faraday's law, induced EMF in primary $E_p = -N_p \frac{d\phi}{dt}$.
Induced EMF in secondary $E_s = -N_s \frac{d\phi}{dt}$.
Dividing the two equations gives the transformer ratio: $\frac{E_s}{E_p} = \frac{N_s}{N_p}$.
For an ideal transformer, input power equals output power: $E_p I_p = E_s I_s \implies \frac{E_s}{E_p} = \frac{I_p}{I_s}$.

Solution:
Interference Conditions (YDSE):
Let two coherent waves from slits $S_1, S_2$ with amplitudes $a_1, a_2$ and phase difference $\phi$ superpose at a point P. The resultant intensity $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$.
Constructive Interference (Maxima): Intensity is maximum when $\cos\phi = +1$.
Phase difference $\phi = 2n\pi$ ($n = 0, 1, 2...$).
Path difference $\Delta x = \frac{\lambda}{2\pi}\phi = n\lambda$.
Destructive Interference (Minima): Intensity is minimum when $\cos\phi = -1$.
Phase difference $\phi = (2n-1)\pi$ ($n = 1, 2...$).
Path difference $\Delta x = \frac{\lambda}{2\pi}\phi = (2n-1)\frac{\lambda}{2}$. Solution (Alternative):
Brewster's Law: When unpolarised light strikes a transparent medium at a specific polarizing angle ($i_p$), the reflected light is completely plane-polarized. The refractive index of the medium is equal to the tangent of the polarizing angle: $\mu = \tan(i_p)$.
Proof of 90°:
By Snell's Law, $\mu = \frac{\sin i_p}{\sin r}$.
By Brewster's Law, $\mu = \tan i_p = \frac{\sin i_p}{\cos i_p}$.
Equating both: $\frac{\sin i_p}{\sin r} = \frac{\sin i_p}{\cos i_p} \implies \sin r = \cos i_p$.
Since $\cos i_p = \sin(90^\circ - i_p)$, we get $r = 90^\circ - i_p \implies i_p + r = 90^\circ$.
The angle between the reflected and refracted rays is $180^\circ - (i_p + r) = 180^\circ - 90^\circ = 90^\circ$. Hence proved.